only due to sets that contain both Maker's element and Breaker's element. These sets gain but then lose , so all in all they lose . Since such sets have at least two elements, each such set loses at most 1/4. By the limited-pair-degree assumption, the number of such sets is at most d2. Hence, after each round the potential-sum drops by at most d2/4. The number of rounds is |X|/2, so the final potential is smaller than the initial potential by at most . The initial potential is . If , the final potential is more than 0, so there is at least one winning-set with potential 1. This set is owned by Maker. Chromatic numbers and winning strategies The chromatic number of is the smallest number of colors needed to color the elements of X such that no single set in is monochromatic. If the chromatic number of is 3, then Maker has a winning-strategy. Summary table The following table summarizes some conditions that guarantee that one of the players has a winning strategy. The condition in the "tightness" column indicates when specific hypergraphs are known on which the strategy stops working. In all conditions, k is the size of winning-sets (i.e., the game hypergraph is k-uniform). Breaker-Breaker game It is possible to play a game in which both players want to attain the goal of Breaker (i.e., have at least one element in each winning-set). Then, the game is not necessarily zero-sum - it is possible that both players will win. In fact, whenever Breaker has a winning strategy in the Maker-Breaker game, it is possible that two Breakers will both win in the Breaker-Breaker game. An application of this strategy is an efficient algorithm for coloring a hypergraph. Suppose we want to color the vertices of a k-uniform hypergraph in two colors such that in each hyperedge, both colors are represented. Erdos proved already in 1963, using the probabilistic method, that such a coloring exists whenever the number of hyperedges is less than (see Property B). However, the proof was not constructive. Using Breaker's constructive winning strategy, we can color the hypergraph by letting two Breakers play one against the other with their winning strategies. Both players will win - so each player will have at least one vertex in every hyperedge. Partial making Suppose that, in order to win, Maker does not need to occupy